| Step | Formula / Calculation | Reasoning |
|---|---|---|
| 1 | \(F_{\text{net}} = m \cdot a\) | Net force calculation using Newton’s second law |
| 2 | \(F_{\text{horizontal}} = F_{\text{net}}\) | Horizontal force component causing acceleration |
| 3 | \(F_{\text{vertical}} = \sqrt{F^2 – F_{\text{horizontal}}^2}\) | Vertical force component calculation |
| 4 | \(\tan(\theta) = \frac{F_{\text{vertical}}}{F_{\text{horizontal}}}\) | Relationship between force components and angle |
| 5 | \(\theta = \arctan\left(\frac{F_{\text{vertical}}}{F_{\text{horizontal}}}\right)\) | Calculating the angle in radians |
| 6 | \(\theta_{\text{degrees}} = \frac{\theta \cdot 180}{\pi}\) | Converting angle to degrees |
| 7 | 68.21° | Final Answer |
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While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
A block of mass \(m\) is accelerated across a rough surface by a force of magnitude \(F\) exerted at an angle \(\theta\) above the horizontal. The frictional force between the block and surface is \(f\). Find the acceleration of the block (as an equation).
An object weighs \( 300 \, \text{N} \) on Earth and \( 50 \, \text{N} \) on the Moon. Does the object have less inertia on the Moon?
The International Space Station has a mass of \(4.2 \times 10^{5} \, \text{kg}\) and orbits Earth at a distance of \(4.0 \times 10^{2} \, \text{km}\) above the surface. Earth has a radius of \(6.37 \times 10^{6} \, \text{m}\) and a mass of \(5.97 \times 10^{24} \, \text{kg}\). Calculate the following:
A ring of negligible mass remains in static equilibrium under the influence of three coplanar forces, as shown in the accompanying diagram. Two forces, each with a magnitude of \(10 \, \text{N}\), act on the ring at an angle \(\theta\) above the horizontal—one directed to the right and the other to the left. A third force, \(F\), acts vertically downward. Determine the magnitude of force \(F\).
In the diagram shown, a \(20 \, \text{N}\) force is applied to block \(B\) (\(7 \, \text{kg}\)). Block \(A\) has a mass of \(3 \, \text{kg}\). Assume frictionless conditions.
A \( 1000 \) \( \text{kg} \) car is traveling east at \( 20 \) \( \text{m/s} \) when it collides perfectly inelastically with a northbound \( 2000 \) \( \text{kg} \) car traveling at \( 15 \) \( \text{m/s} \). If the coefficient of kinetic friction is \( 0.9 \), how far, and at what angle do the two cars skid before coming to a stop?
Determine the force needed to push a \( 150 \) \( \text{kg} \) body up a smooth \( 30^\circ \) incline with an acceleration of \( 6 \) \( \text{m/s}^2 \).
Shown above are three masses of \(6 \, \text{kg}\), \(3 \, \text{kg}\), and \(1 \, \text{kg}\) (in order from left to right). You pull on the 1kg mass with a force \(F\) of \(15 \, \text{N}\) along a frictionless surface.
68.2° above the horizon
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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